$\begingroup$1) isn't true because we can get $\alpha\beta = a$ (a algebraic) by simply setting $\beta = a/\alpha$. 4) is true because if $\alpha^2$ solves $P(x)$ than $\alpha$ solves $P(x)^2$. 2) is true because... 3) isn't true because for any algebraic $a$ and any transendental $b$ $a= b^{\log_b a}$ and $\log_b a$ is trans. (can replace $b$ with $e$ if you want.)$\endgroup$
– fleabloodFeb 10 '16 at 6:48

$\begingroup$Regarding 2): In general, you can think of any algebraic dependencies that $\alpha$ has over a ground field $K$ as obstructions to $K(\alpha) \simeq K(X)$, where $X$ is an indeterminate.$\endgroup$
– mad_algebraistFeb 10 '16 at 6:57

Notice that $e$ and $\ln 2$ are transcendental, but $e^{\ln 2} = 2$. So this is false.

If $\alpha$ is transcendental, then $\alpha^n$ is transcendental for every $n \in \mathbb N$ (if there were a nontrivial polynomial $p(x) \in \mathbb Z[x]$ with $p(\alpha^n) = 0$, then let $q(x)$ be the polynomial obtained from $p(x)$ by replacing every $x^k$ in $p(x)$ with $x^{n\cdot k}$. Then $q(x)$ is nontrivial and $q(\alpha) = p(\alpha^n) = 0$, so $\alpha$ is algebraic - Contradiction!)

$\begingroup$$\ln 2$ is transcendental by the Hermite-Lindemann Transcendence Theorem, which is difficult. There do exist trans. numbers $a,b$ with $a^b$ trans., because the algebraic numbers form a countable set, whereas the set of reals is uncountable.$\endgroup$
– DanielWainfleetFeb 10 '16 at 6:57

$\begingroup$@user254665 I never claimed that $\alpha^\beta$ couldn't be transcendental for transcendental $\alpha$ and $\beta$. Instead I provided an example for transcendentals $\alpha$ and $\beta$ such that $\alpha^\beta$ is algebraic (or in my case - an integer).$\endgroup$
– Stefan MeskenFeb 10 '16 at 7:01

$\begingroup$.No of course you didn't. I was adding something else.$\endgroup$
– DanielWainfleetFeb 10 '16 at 13:01

Here is a proof that point 3 is false that doesn't use any difficult theorems.

As $\gamma$ ranges through all transcendental numbers, the number $2^\gamma$ takes on uncountably many values. Since only countably many of these values can be algebraic, there must exist a transcendental $\gamma$ for which $2^{\gamma}$ is also transcendental.